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Homework 8
Directions:
Show each step of your work and fully simplify each expression.
Turn in your answers in class on a physical piece of paper.
Staple multiple sheets together.
Feel free to use Desmos for graphing.
Answer the following:
If \[f(x) = x^2 - 1 \qquad g(x) = x - 1\] find the function $\dfrac{f(x)}{g(x)}$ and state it's domain.
Redefine these rational functions with their domain exclusions (also called
holes
)
$f(x) = \dfrac{x^2 - 3x + 2}{x - 1}$
$g(t) = \dfrac{2t^2 - 5t + 2}{2t - 1}$
$f(x) = \dfrac{(x-3)(x^2 - 2x + 1)}{x-1}$
Find the range of the following functions. Use transformations!
$f(x) = \sqrt{x}$
$f(x) = \sqrt{x - 1}$
$f(x) = -\sqrt{x + 1}$
$f(x) = 4 - x^2$
Draw a graph of a one-to-one function.
Suppose $f$ is
not
a one-to-one function. Explain what property is violated when we try to define the inverse $f^{-1}$.
Determine if the following functions are inverses of each other using the Inverse Function Property.
$f(x) = 2 - 5x, \ g(x) = \dfrac{2 - x}{5}$
$f(x) = x^2, \ g(x) = \sqrt{x}$
$f(x) = \dfrac{1}{x - 1}, \ g(x) = \dfrac{1}{x} + 1$
$f(x) = x + 1, \ g(x) = \dfrac{1}{x - 1}$
The function $f(x) = (x-1)^2$ is not one-to-one. What could we restrict the domain to to make $f(x)$ one-to-one?
Given this graph of a function:
find
$f^{-1}(2)$
$f^{-1}(5)$
$f^{-1}(6)$
If $f(1) = 2, f(2) = 3$ and $f(3) = 1$ find $f^{-1}(1), f^{-1}(2), f^{-1}(3)$.
Find the inverse function of each of the following functions:
$f(x) = 3x + 5$
$f(x) = -2x - 4$
$f(x) = \dfrac{1}{x + 2}$
$f(x) = \dfrac{2x + 5}{x - 7}$
$f(x) = 4-x^2, \qquad x \geq 0$
$f(x) = \sqrt{4 - x^2}, \qquad 0 \leq x \leq 2$
$f(x) = \dfrac{1}{x^2}, \qquad x > 0$
Why is it helpful to put a quadratic into standard form?
A quadratic $f(x) = x^2 + bx$ is given. What do you need to add and subtract in order to complete the square?
A quadratic function $f(x)$ is given. Do the following four tasks for each function:
Express $f(x)$ in standard form.
Sketch a graph of $f(x)$.
(skip this) Find the maximum or minimum value of $f(x)$.
(skip this) Describe the type of solutions of $f(x)$ (use the discriminant).
$x^2 - 2x + 3$
$-x^2 + 6x + 4$
Show Hint
Rewrite $-x^2 + 6x + 4 = -x^2 - (-6)x + 4 = -(x^2 - 6x) + 4$ and complete the square.
$-3x^2 + 6x - 2$
$2x^2 + 12x + 10$
$x^2 + 2x$
$x^2 + 4x - 1$